PAT : 03-树2. Tree Traversals Again (25)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

看完题就觉得这道题可以不用根据输入去建个树然后再后续遍历，但是我一开始就很在意题里说的一句话“a unique binary tree can be generated from this sequence of operations”。不是说要一个中序加上一个先序或者后序才能唯一确定一个树吗？但给出的操作只告诉了我们中序啊？看来输入一定不只有中序的信息，想了半天才想到输入的push的顺序不就是先序序列吗，于是根据这两个序列，后序序列就出来了，我用的是递归的方法做的。

#include <stdio.h>
#include <stdlib.h>

int n;

void genPostOder (int preOrder[], int inOrder[], int size)
{
	if(size == 1 && preOrder[0] == inOrder[0])
		printf("%d ", preOrder[0]);

	else if(size > 1)
	{
		int length1, length2;
		int root = preOrder[0];
		int i;
		for(i = 0; i < size; i++)
			if(inOrder[i] == root)
				break;
		int rootp = i;
		length1 = rootp;
		length2 = size - rootp - 1;

		int *inOrder1 = (int *) malloc (length1 * sizeof(int));
		int *inOrder2 = (int *) malloc (length2 * sizeof(int));
		int l1=0, l2=0;
	
		for(i = 0; i < size; i++)
		{
			if(i < rootp)
				inOrder1[l1++] = inOrder[i];
			else if(i > rootp)
				inOrder2[l2++] = inOrder[i];
		}
		int *preOrder1 = (int *) malloc (length1 * sizeof(int));
		int *preOrder2 = (int *) malloc (length2 * sizeof(int));
		l1=0, l2=0;

		for(i = 1; i < size; i++)
		{
			if(i <= length1)
				preOrder1[l1++] = preOrder[i];
			else
				preOrder2[l2++] = preOrder[i];
		}

		genPostOder(preOrder1, inOrder1, length1);
		genPostOder(preOrder2, inOrder2, length2);
		if(size < n)
			printf("%d ", root);
	}
}

void push (int stack[], int *length, int num)
{
	stack[(*length)++] = num;
}

int pop (int stack[], int *length)
{
	return stack[--(*length)];
}

int main()
{
	char ch;
	int node;
	scanf("%d", &n);
	getchar();
	int *inOrder = (int *) malloc (n * sizeof(int));
	int *preOrder = (int *) malloc (n * sizeof(int));
	int *stack = (int *) malloc (n * sizeof(int));
	int length=0, inl=0, prel=0;
	int i;
	for(i = 0; i < 2*n; i++)
	{
		scanf("P%c", &ch);
		if(ch == 'u')
		{
			scanf("sh %d", &node);
			getchar();
			push(stack, &length, node);
			preOrder[prel++] = node;
		}
		else if(ch == 'o')
		{
			scanf("p");
			getchar();
			inOrder[inl++] = pop(stack, &length);
		}
	}

	if(n > 1)
		genPostOder(preOrder, inOrder, n);
	printf("%d\n", preOrder[0]);

	return 0;
}